Julv 1994
LIDS-P-2254
RESIDUAL NOISE AFTER STRIPPING ON
FADING MULTIPATH CHANNELS
R. G. Gallager, M.I.T. 5/9/94
1) INTRODUCTION: We are interested in transmitting pseudo-noise (CDMA)
signals over a multipath channel, measuring the channel from the output, and using that
measurement to subtract the effect of the transmitted signal from the received waveform.
The reason for wanting to do this is to take advantage of stripping for multiaccess
communication, but we can study the channel measurement and the noise introduced by
stripping without reference to the multiple sources. Thus here we look at only a single
source.
LOW PASS EQUIVALENTS: Consider M-ary signalling with the M signals
ul(t),...,uM(t). Let T be the intersymbol duration, so that each T seconds, one of the
signals {um(t), lm•}MI is transmitted. We assume that um(t) is essentially non-zero only
for O<t<T so that successive signals do not overlap. After passing through the multipath
channel there will be some overlap which we discuss later. The signals all have bandwidth
W, centered around some carrier frequency fo>>W. Letting Um(f) = I um(t)e-j 2n ft dt be
the Fourier transform of um(t) for each m, we take the low pass equivalent waveforms
xm(t) with Fourier transforms Xm(f) to be defined by Xm(f) = Um(f+f 0), f>-fo and Xm(f)
= 0 otherwise (see Fig. 1).
-fo
ri-.
Ito
0
, -,
Xm(f)
fo
Um(f)
f
[
_-
W
freq
shift
|
Figure 1: Low pass equivalent; Um(f) and Xm(f) are actually complex.
This particular way of going from passband to baseband is not entirely conventional. It has
the peculiar aspect that
um(t) = 2Re[xm(t)ej2fot] = 2Re[xm(t)]cos(2'fot) - 2Im[xm(t)]sin(2i7fot)
An even more peculiar aspect is that if E is the energy in the signal um(t), so that
(1)
J
[uM(t)]
2
Um(f)l 2 df = E
[Um(t)] 2 dt = |
dt =
(2)
then the low pass waveform (since it is a frequency translate only of the positive frequency
part of u) contains only half the signal energy E:
Ix(t)12 dt =
f
IXm(f) 2 df = E2
2,0~~~~~~
~(3)
The desirable feature of this scaling is that if u(t) is passed through a linear filter of impulse
response g(t) to get output v(t), and if x(t), h(t), and y(t) are the corresponding low pass
equivalents of u(t), g(t), and v(t), then
v(t) = u(t)*g(t); V(f) = G(f)V(f);
y(t) = h(t)*x(t); Y(f) = H(f)X(f)
(4)
Since the arguments to follow depend critically on being able to view signals and filters
interchangeably, we have defined low pass waveforms so that (4) is satisfied, and by
necessity this forces the peculiar energy scaling in (3).
The particular signals used in multiaccess systems are usually pseudo-noise signals (often
called CDMA signals). They have the property that lUm(f)l is essentially constant over the
signalling bandwidth If-fol < W/2. It follows that IXm(f)l is essentially constant over Ifl <
W/2 and 0 elsewhere. For simplicity, we henceforth assume that IXm(f)l is exactly constant
over IFI<W/2 and zero elsewhere. Using (3) then,
IXm(f)12 =2E- for Ifi < W/2; IXm(f)12 = 0 for Ifl > W/2
2W
(5)
Since IXm(f)12 and Rm('r) = | xm*(t)Xm(t+r) dr are Fourier transforms, it follows from (5)
that Rm(t) has a (sin x)/x form. Thus, if we view Xm(t) in terms of its samples at rate W,
we have Yi X*m,i Xm,i+j = (E W/2) 5(j). Eq.(5) appears to be more intuitively satisfying
for those who don't work with sampled data systems constantly, so we use that in what
follows.
It is not possible to find waveforms x(t) that are both time limited to the signal interval T
and low pass limited to the band W/2. CDMA systems, however, have a relatively large
time bandwidth product, WT >> 1 (which is why they are called spread spectrum systems),
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and for this reason, waveforms can be found that are both approximately time limited and
frequency limited. Finding such waveforms with desirable cross correlation properties is a
large and very well studied field, but studying this would draw us away from our main
purpose. Thus in what follows, we simply assume (5) to be valid, and recognize that the
approximation can be quite good for WT >> 1.
THE EFFECT OF MULTIPATH: Let ai(t) be the strength of the ith propagation path
at time t, and let xi(t) be the propagation delay of that path. Both ai and ri change slowly
with time. The impulse response of the channel, i.e., the output at time t due to an impulse
t seconds earlier is then g(r,t) = Ai ai(t) 8(t-ti(t)). Thus the response to a signal u(t) is
v(t) = f u(t-t)g(t,t)dT = Yi u(t-Ti(t))ai(t)
(6)
Defining G(f,t) = I g(t,t)e-j 2 fz fX dt, we have G(f,t) = Hi ai(t) e-j2nfxi(t). Since the input to
the channel is bandlimited, we are only interested in G(f,t) in the band If-fol < W/2. Thus
we define H(f,t) = G(f+fo,t) for Ifl<W/2. Thus, for Ifl < W/2,
H(f,t) = Ei ai(t) e-j27f0xi(t)e-j2tfxi(t) = ijxai(t)e-j2 xfti(t) where ci(t) = ai(t)e-j2 nfoxi(t)
Inverse Fourier transforming, the low pass equivalent filter, for Ifl < W/2, is
h(x,t) = H'i ai(t) sin[rcW(t-ti(t))]
(7)
C(t-xi(t))
Note that h(t,t) has one filtered impulse for each path, and that the sin x over x pulse
representing the filtered impulse has a peak that increases with W and a width that decreases,
thus keeping unit area. The multipath structure does not change as the bandwidth of the input
is changed, but the filter h(t,t) does change, since h(t,t) represents only the effect of the
channel over the given bandwidth. This is an important point, since the effect of the channel
is typically very complex, but we need measure its effect only on the signals in the given
band. Since we want to measure the channel over the bandwidth W, we want to characterize
it in the simplest way over that band.
Define L as the multipath spread of the channel; this is the difference in propagation delay
between the longest and shortest path. Thus h(r,t) is approximately 0 for t<0 and for v>L
(in communication, one usually adjusts the time reference at the receiver to be delayed from
that at the transmitter by the shortest (or sometimes the strongest) propagation delay). For
cellular mobile communication, L is typically between a few microseconds and 30
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microseconds, and for the personal communication systems of the future, L is typically
much smaller, on the order of 100 nsec. If L = 10 gsec, and W = 106 H, then h(t,t) could
be represented (through the sampling theorem) by 10 samples in t; each sample is complex,
so measuring h at a given t corresponds to measuring 20 real numbers.
Define B as the Doppler spread of the channel; this is the difference between the largest and
the smallest Doppler shift. Typical values in a mobile system are around 100 H. B
determines how quickly h(x,t) can change with t. The phases in the path strengths cXi(t) can
change significantly over the interval 1/B, so that measurements of the channel become
outdated over intervals of duration 1/B. We will assume in what follows that the signalling
interval T is very much smaller than 1/B, and thus we assume that h(t,t) is constant as a
function of t over a signal interval T. Thus h(t,t) is modeled as a linear time invariant filter
over individual signal intervals, allowing one to play all the games of elementary linear
systems. One must recognize, of course, that h(r,t) changes significantly over many
signalling intervals, so that one cannot simply measure h once and for all.
ESTIMATING h(t,t): First ignore noise, assume that xm(t) is transmitted, and consider
passing the channel output, xm(t)*h(t,t) through a filter matched to Xm (i.e., a filter with
impulse response x*m(-z)).
vm(t)
Xm(Z)
X*m(-)
Eh(~,t)
Figure 2
Taking Fourier transforms, we have
Vm(f) = Xm(f) H(f,t) X*m(f) = IXm(f)12 H(f,t) =
H(f,t)
where we have used (5). Taking the inverse Fourier transform, vm(t)
=-W
(8)
h(t,t). Since
we are looking at an input in the interval (0,T), and we are assuming that h(t,t) does not
change in t over intervals of duration T, the parameter t can be taken to be 0. This
suggests that the output should be attenuated by 2W/E in order to obtail ian estimate of
htt).
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We now put the white noise back in the picture and look at the output of the attenuated
matched filter including noise (see Fig 3). Assume the noise has spectral density N0/2.
Filtering the noise to If-fol < W/2, and defining the base band equivalent noise, as before,
as the upper sideband shifted down by fo, the baseband equivalent noise process is
Gaussian and has the spectral density No/2 for Ifl < W/2. It follows that the noise power of
the baseband waveform is NoW/2, which is half the noise power of the band pass
waveform. Thus we have scaled the noise in the same way as the signal. Physically,
when one demodulates a passband waveform into quadrature baseband components, one
can scale those baseband waveforms arbitrarily, but the signal and noise must be scaled the
same way.
Xm(:)
} h(,t)
Xm(-:)'2W-
h(-1t) +Zn(F)g
Figure 3
We have seen that the component of the output due to the signal xm(t) is h(',t). To
analyze a sample Zm(t) of the output noise process, note that the Fourier transform of the
attenuated matched filter is X*m(f)2W/E. The spectral density of the filter, i.e., the
magnitude squared of the Fourier transform, is (E /(2W))[2W/E ]2 = 2WIE . Since the
input process has the spectral density No/2, the output process {Zm(T:)} has the spectral
density NoW/E . Thus the noise power of {Zm(t) } is NoW 2 /E and
E[Zm(t) 2 ] = NoW2 /E.
(9)
Now consider a rake receiver (see Figure 4). Assume that h(r,t) has been well estimated,
and thus that the output of the filter matched to h(t,t) allows a correct decision to be made
on m. Given this decision, the output of the mth filter matched to the signal yields a new
estimate of h plus additive Gaussian noise. The device to estimate h then uses the decision
on m to accept the input from the m th filter to update the old estimate of h.
To avoid worrying about the optimal estimate of h, we can get a good approximate answer
by simply assuming that the estimate of h is simply the linear average of the previous n
measurements (where each measurement comes from appropriate m). Since the filter
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x*1(-rl) 2W
h('.)1(
Xm00
rc,t)+Zm(,
h,(-:,t)
2W
,--
h(\,t)
Estimate
eid
k
Figure 4: Rake Receiver
remains almost constant for a time on the order of 1/B, we take n = 1/(BT). Since taking
an average over n measurements reduces the noise variance by a factor of n, we have
h(t,t) = h(c,t) + z'(t) ;
2
E[(Z'(t))2 ] = N W -_-BTNoW
nE
E
O_<_L
(10)
where z'(t) is a sample function of the Gaussian random process {Z'(t) }. We assume that
the multipath spread L is known, and thus that h(t,t) is taken to be 0 for t<0 and t>L.
Since all of the noise processes being averaged are white over Ifl < W/2, {Z'(t) I is also
white over Ifl_<W/2.
RESIDUAL NOISE: Finally we have the problem of determining the residual noise if
the effect of the detected signal is subtracted from the received waveform (again assuming
the correct signal was detected). The effect of the signal xm(t) on the received signal is
Xm(T)*h(t,t). The quantity subtracted from the received signal in stripping is xm(t)*h(x,t),
and thus the residual noise after stripping is ¢(c) = xm(T)*z'(T). Taking Fourier
transforms,
(f) = Xm(f)Z'(f);
Thus,
I
= tXm(f)121Z'(f) 2
=(f)12
Ew IZ'(f)i 2
T2W
(11)
7
J kI(f)12 df= J fIZ'(f)12df = J 1O(()1 2 d' = J Z,('r)I2 dt
2W
2W
(12)
Taking the expected value of both sides of (12),
E[10()I
2 ']d l
=
E
LJ()t
J
E[IZ'(,)12]d: = BTNOWL
L.2Wt2
(13)
This is the baseband expected energy of the residual noise in the band Ifl<W and over the
interval O<t•T. Since Z'(t) is white over the band IfltW/2, ¢(X) is also white over Ifll-W.
Thus the spectral density of this noise power (averaged over the time interval (O,T)) is
BLNO/2. Since the spreading product BL is small for most wireless situations, this
indicates that the residual noise is small relative to the ordinary additive noise of spectral
density No/2.
When multiaccess communication is taken into account, the noise that effects the filter
measurement becomes not only the white noise but also the other users signals, which have
been passed through their own multipath filters before contributing to the measurement of
the filter in question.
The argument here does not necessarily mean that stripping will work. It has assumed that
decisions can be made correctly, and has not adjusted for the effect of incorrect decisions.
The argument has also assumed that correct decisions can be made without delay (which is
not necessarily correct in coded systems). Finally, perfect timing recovery has been
assumed. What we have shown, however, is that there is no inherent reason why stripping
can not work on multipath channels.